DPP for NEET: Daily Practice Problems, Ch 52: Wave Optics- 1 (Solutions)

# Wave Optics- 1 Practice Questions - DPP for NEET

``` Page 1

1. (d)
1
2
I 9
I1
=
2
1
2
2 max
min 1
2
I
1
I I 91
I I 91
1
I
éù
+
êú
éù
+
êú
==
êú
êú
-
ëû
êú-
êú
ëû
2
max
2
min
I 44
I1
2
==
2. (a)a
1
= 6 units, a
2
= 8 units
2
2
1
2 max
22
min
1
2
a
6
1
1
a I 8
I
6
a
1
1
8
a
éù
éù
+
+ êú
êú
ë û ëû
==
é ù éù
-
-
ê ú êú
ëû
ëû
max
min
I 49
I1
=
3. (b) The separation between the sucessive bright fringes
is-
b =
9
3
D 1 600 10
d
0.1 10
-
-
l ´´
=
´
b = 6.0 mm.
4. (b) w
a
= l/d
\ w
a
µ l  Þ
a water
a
() w
w
=
water
l
l
Þ
a water
a
() w
w
=
water
l
ml
Þ (w
a
)
water
= 0.15°
5. (b) I' = I
1
+ I
2
+ 2
12
I I cosf
I
1
= I, I
2
= 9I, f = p
I' = I + 9I + 2
2
9I cos p = 10I – 6I = 4I
6. (a) d =
Dl
b
.... (1)
According to quesion.
l = 5100 × 10
-10
m
b =
2
10
× 10
–2
m .... (2)
D = 2m,  d = ?.
From eqs. (1) and (2)
d =
8
3
2 51 10
2 10
-
-
´´
´
= 5.1 × 10
-4
m
7. (b) V =
max
max min min
max
max min
min
I
1
II I
I
II
1
I
-
-
=
+
+
.... (1)
2
1
2 max
min 1
2
I
1
I I
I I
1
I
æö
+
ç÷
ç÷
=
ç÷
ç÷ -
ç÷
èø
.... (2)
According to question
1
2
I 1
I4
=
.... (3)
From eqs. (2) and (3)
2
m ax
m in
1 9
1
I
4 4
9
1
I
1
1
4
4
æö
+
ç÷
ç÷
= ==
ç÷
-
ç÷
èø
.... (4)
From eqs. (1) and (4)
V =
[9 1]
[9 1]
-
+
=
8
10
= 0.8
8. (c) l =
6
( 1)t (1.6 1) 7 10
n7
-
m- - ´´
=
l = 6 × 10
–7
meter Þ l = 6000 Å
9. (d) b µ
1
d
\ d  On increasing d three times
b will become 1/3 times.
10. (b) \ PR = d Þ PO = d secq and CO = PO cos 2q
= d sec q cos 2q is
Q
O
C
A
P
B
R
q q
q
Page 2

1. (d)
1
2
I 9
I1
=
2
1
2
2 max
min 1
2
I
1
I I 91
I I 91
1
I
éù
+
êú
éù
+
êú
==
êú
êú
-
ëû
êú-
êú
ëû
2
max
2
min
I 44
I1
2
==
2. (a)a
1
= 6 units, a
2
= 8 units
2
2
1
2 max
22
min
1
2
a
6
1
1
a I 8
I
6
a
1
1
8
a
éù
éù
+
+ êú
êú
ë û ëû
==
é ù éù
-
-
ê ú êú
ëû
ëû
max
min
I 49
I1
=
3. (b) The separation between the sucessive bright fringes
is-
b =
9
3
D 1 600 10
d
0.1 10
-
-
l ´´
=
´
b = 6.0 mm.
4. (b) w
a
= l/d
\ w
a
µ l  Þ
a water
a
() w
w
=
water
l
l
Þ
a water
a
() w
w
=
water
l
ml
Þ (w
a
)
water
= 0.15°
5. (b) I' = I
1
+ I
2
+ 2
12
I I cosf
I
1
= I, I
2
= 9I, f = p
I' = I + 9I + 2
2
9I cos p = 10I – 6I = 4I
6. (a) d =
Dl
b
.... (1)
According to quesion.
l = 5100 × 10
-10
m
b =
2
10
× 10
–2
m .... (2)
D = 2m,  d = ?.
From eqs. (1) and (2)
d =
8
3
2 51 10
2 10
-
-
´´
´
= 5.1 × 10
-4
m
7. (b) V =
max
max min min
max
max min
min
I
1
II I
I
II
1
I
-
-
=
+
+
.... (1)
2
1
2 max
min 1
2
I
1
I I
I I
1
I
æö
+
ç÷
ç÷
=
ç÷
ç÷ -
ç÷
èø
.... (2)
According to question
1
2
I 1
I4
=
.... (3)
From eqs. (2) and (3)
2
m ax
m in
1 9
1
I
4 4
9
1
I
1
1
4
4
æö
+
ç÷
ç÷
= ==
ç÷
-
ç÷
èø
.... (4)
From eqs. (1) and (4)
V =
[9 1]
[9 1]
-
+
=
8
10
= 0.8
8. (c) l =
6
( 1)t (1.6 1) 7 10
n7
-
m- - ´´
=
l = 6 × 10
–7
meter Þ l = 6000 Å
9. (d) b µ
1
d
\ d  On increasing d three times
b will become 1/3 times.
10. (b) \ PR = d Þ PO = d secq and CO = PO cos 2q
= d sec q cos 2q is
Q
O
C
A
P
B
R
q q
q
DPP/ P 52
146
Path difference between the two rays
D = CO + PO = (d sec q + d secq cos 2q)
Phase difference between the two rays is
f = p (One is reflected, while another is direct)
Therefore condition for constructive interference
should be
D =
3
,
22
ll
.......
or d sec q(1 + cos 2q) =
2
l
or ( )
2
d
2 cos
cos
q
q
=
2
l
Þ cos q =
4d
l
11. (b) n
1
l
1
= n
2
l
2
10 × 7000 = n
2
× 5000
n
2
= 14
12. (a, c) Path difference between the rays reaching infront of
slit S
1
is.
S
1
P – S
2
P = (b
2
+ d
2
)
1/2
– d
For distructive interference at P
b
S
1
S
2
d
P
S
1
P – S
2
P =
( ) 21
2
n-l
i.e., (b
2
+ d
2
)
1/2
– d =
( ) 21
2
n-l
Þ
( )
1/2
2
2
2n1
b
d1d
2
d
æö
-l
+ -= ç÷
ç÷
èø
Þ
( )
2
2
2n1 b
d 1 ...... d
2
2d
æö
-l
+ + -= ç÷
ç÷
èø
(Binomial Expansion)
Þ
( ) 2n1 b
2d2
-l
=
Þ =
( )
2
b
2n 1d -
For n = 1,2 ............, l =
22
bb
,
d 3d
13. (b) Distance of mth bright fringe from central fringes is
X
m
=
mD
d
l
14. (a, b) For microwave l =
c
f
=
8
6
3×10
10
= 300 m
d
Y
D
P
q
S
1
S
2
q
Dx
As Dx = d sin q
Phase difference f =
2p
l
(Path difference)
= ( )
2
d sin
p
q
l
=
2
300
p
(150 sin q) = p sin q
I
R
= I
1
+ I
2
+
12
2 I I cosq
Here I
1
= I
2
and f = p sin q
\ I
R
= 2I
1
[(p sin q)] = 4I
1
cos
2
sin
2
pq æö
ç÷
èø
I
R
will be maximum when cos
2

sin
2
pq æö
ç÷
èø
= 1
\ (I
R
)
max
= 4I
1
= I
o
Hence I = I
o
cos
2

sin
2
pq æö
ç÷
èø
If q = 0, then I = I
o
cosq = Io
If q = 30°, then I = I
o
cos
2
(p/4) = I
o
/2
If q = 90°, then I = I
o
cos
2
(p/2) = 0
15. (b) I = R
2
= a
1
2
+ a
2
2
+ 2a
1
a
2
cos d
=  I + 4I + 4I cos
2
p
= 5I
16. (b) Suppose the amplitude of the light wave coming from
the narrower slit is A and that coming from the wider
slit is 2A. The maximum intensity occurs at a place
where constructive interference takes place. Then the
resultant amplitude is the sum of the individual
amplitudes.
Thus,
A
max
= 2A + A = 3A
The minimum intensity occurs at a place where
destructive interference takes place. The resultant
amplitude is then difference of the individual
amplitudes.
Thus, A
min
= 2A – A = A.
\
2 2
max max
22
min
min
I (A) (3A)
9
I
(A ) (A)
= ==
Page 3

1. (d)
1
2
I 9
I1
=
2
1
2
2 max
min 1
2
I
1
I I 91
I I 91
1
I
éù
+
êú
éù
+
êú
==
êú
êú
-
ëû
êú-
êú
ëû
2
max
2
min
I 44
I1
2
==
2. (a)a
1
= 6 units, a
2
= 8 units
2
2
1
2 max
22
min
1
2
a
6
1
1
a I 8
I
6
a
1
1
8
a
éù
éù
+
+ êú
êú
ë û ëû
==
é ù éù
-
-
ê ú êú
ëû
ëû
max
min
I 49
I1
=
3. (b) The separation between the sucessive bright fringes
is-
b =
9
3
D 1 600 10
d
0.1 10
-
-
l ´´
=
´
b = 6.0 mm.
4. (b) w
a
= l/d
\ w
a
µ l  Þ
a water
a
() w
w
=
water
l
l
Þ
a water
a
() w
w
=
water
l
ml
Þ (w
a
)
water
= 0.15°
5. (b) I' = I
1
+ I
2
+ 2
12
I I cosf
I
1
= I, I
2
= 9I, f = p
I' = I + 9I + 2
2
9I cos p = 10I – 6I = 4I
6. (a) d =
Dl
b
.... (1)
According to quesion.
l = 5100 × 10
-10
m
b =
2
10
× 10
–2
m .... (2)
D = 2m,  d = ?.
From eqs. (1) and (2)
d =
8
3
2 51 10
2 10
-
-
´´
´
= 5.1 × 10
-4
m
7. (b) V =
max
max min min
max
max min
min
I
1
II I
I
II
1
I
-
-
=
+
+
.... (1)
2
1
2 max
min 1
2
I
1
I I
I I
1
I
æö
+
ç÷
ç÷
=
ç÷
ç÷ -
ç÷
èø
.... (2)
According to question
1
2
I 1
I4
=
.... (3)
From eqs. (2) and (3)
2
m ax
m in
1 9
1
I
4 4
9
1
I
1
1
4
4
æö
+
ç÷
ç÷
= ==
ç÷
-
ç÷
èø
.... (4)
From eqs. (1) and (4)
V =
[9 1]
[9 1]
-
+
=
8
10
= 0.8
8. (c) l =
6
( 1)t (1.6 1) 7 10
n7
-
m- - ´´
=
l = 6 × 10
–7
meter Þ l = 6000 Å
9. (d) b µ
1
d
\ d  On increasing d three times
b will become 1/3 times.
10. (b) \ PR = d Þ PO = d secq and CO = PO cos 2q
= d sec q cos 2q is
Q
O
C
A
P
B
R
q q
q
DPP/ P 52
146
Path difference between the two rays
D = CO + PO = (d sec q + d secq cos 2q)
Phase difference between the two rays is
f = p (One is reflected, while another is direct)
Therefore condition for constructive interference
should be
D =
3
,
22
ll
.......
or d sec q(1 + cos 2q) =
2
l
or ( )
2
d
2 cos
cos
q
q
=
2
l
Þ cos q =
4d
l
11. (b) n
1
l
1
= n
2
l
2
10 × 7000 = n
2
× 5000
n
2
= 14
12. (a, c) Path difference between the rays reaching infront of
slit S
1
is.
S
1
P – S
2
P = (b
2
+ d
2
)
1/2
– d
For distructive interference at P
b
S
1
S
2
d
P
S
1
P – S
2
P =
( ) 21
2
n-l
i.e., (b
2
+ d
2
)
1/2
– d =
( ) 21
2
n-l
Þ
( )
1/2
2
2
2n1
b
d1d
2
d
æö
-l
+ -= ç÷
ç÷
èø
Þ
( )
2
2
2n1 b
d 1 ...... d
2
2d
æö
-l
+ + -= ç÷
ç÷
èø
(Binomial Expansion)
Þ
( ) 2n1 b
2d2
-l
=
Þ =
( )
2
b
2n 1d -
For n = 1,2 ............, l =
22
bb
,
d 3d
13. (b) Distance of mth bright fringe from central fringes is
X
m
=
mD
d
l
14. (a, b) For microwave l =
c
f
=
8
6
3×10
10
= 300 m
d
Y
D
P
q
S
1
S
2
q
Dx
As Dx = d sin q
Phase difference f =
2p
l
(Path difference)
= ( )
2
d sin
p
q
l
=
2
300
p
(150 sin q) = p sin q
I
R
= I
1
+ I
2
+
12
2 I I cosq
Here I
1
= I
2
and f = p sin q
\ I
R
= 2I
1
[(p sin q)] = 4I
1
cos
2
sin
2
pq æö
ç÷
èø
I
R
will be maximum when cos
2

sin
2
pq æö
ç÷
èø
= 1
\ (I
R
)
max
= 4I
1
= I
o
Hence I = I
o
cos
2

sin
2
pq æö
ç÷
èø
If q = 0, then I = I
o
cosq = Io
If q = 30°, then I = I
o
cos
2
(p/4) = I
o
/2
If q = 90°, then I = I
o
cos
2
(p/2) = 0
15. (b) I = R
2
= a
1
2
+ a
2
2
+ 2a
1
a
2
cos d
=  I + 4I + 4I cos
2
p
= 5I
16. (b) Suppose the amplitude of the light wave coming from
the narrower slit is A and that coming from the wider
slit is 2A. The maximum intensity occurs at a place
where constructive interference takes place. Then the
resultant amplitude is the sum of the individual
amplitudes.
Thus,
A
max
= 2A + A = 3A
The minimum intensity occurs at a place where
destructive interference takes place. The resultant
amplitude is then difference of the individual
amplitudes.
Thus, A
min
= 2A – A = A.
\
2 2
max max
22
min
min
I (A) (3A)
9
I
(A ) (A)
= ==
DPP/ P 52
147
17. (b)
1
2
I 2
I1
=
11
22
aI 2
a I1
==
At the point of constructive interference, the resultant
amplitude becomes (a
1
+ a
2
) = 2 + 1 at the point of
destructive interference, the resultant amplitude is
(a
1
– a
2
) = 2 – 1
\
2 2
max 12
22
min
12
I (a a) ( 2 1)
34
I
(a a ) ( 2 1)
+ +
= ==
--
18. (d) For destructive interference :
Path difference= S
1
P – S
2
P = (2n –1) l/2.
For  n = 1, S
1
P – S
2
P= (2 × 1 – 1) l/2 = l/2.
n =2, S
1
P– S
2
P = (2 × 2 – 1) l/2 = 3l/2.
n = 3, S
1
P – S
2
P = (2× 3–1) l/2  = 5l/2.
n = 4, S
1
P – S
2
P = (2 × 4 – 1) l/2  = 7l/2.
n = 5,S
1
P – S
2
P = (2 × 5 – 1) l/2= 9l/2.
n =6,S
1
P – S
2
P = (2× 6– 1) l/2  = 11l/2.
So, destructive pattern is possible only for path
difference = 11l/2.
19. (b) The distance of a bright fringe from zero order fringe is
given by-
X
n
=
n D
d
l
D & d is constant
n
1
l
1
= n
2
l
2
n
1
= 16, l
1
= 6000 Aº, l
2
= 4800 Å
n
2
=
11
2
n 16 6000
4800
l ´
=
l
= 20
20. (c)n
1
l
1
= n
2
l
2
for bright fringe
n (7.5 × 10
–5
) = (n + 1) (5 × 10
–5
)
n =
5
5
5.0 10
2.5 10
-
-
´
´
= 2.
21. (b) X
n
=
nD
d
l
or X
3
=
3D
d
l
X
3
=
8
3 (5000 10 cm) 200cm
0.02cm
-
´ ´´
= 1.5 cm.
22. (b, d)
max
max
I
9
I
=
Þ
2
12
12
a + a
a – a
æö
ç÷
èø
= 9  Þ
12
12
a + a
a – a
= 3
Þ
1
2
a
a
=
3 + 1
3 – 1
Þ
1
2
a
a
= 5 There I
1
: I
2
= 4 : 1
23. (a) Fringe width b =
D
d
l
According to de Broglie,
Wavelength l =
hh
p 2meV
=
As V increases, l decreases, b decreases.
Also
1
andD
d
bµ bµ .
24. (a) n =
( 1)tD
d
m-
but b =
D
d
l
Þ
D
d
b
=
l
n = (m – 1) t b/l
20b = (m – 1) 2.5 × 10
–3
{ b/5000 × 10
–8
}
m – 1 =
8
3
20 5000 10
2.5 10
-
-
´´
´
= 0.4
m = 1.4.
25. (a) S
1
P – S
2
P
dy
D
=
For central maxima,
Dx = (n
0
+ kt) –sin0
dy
d
D
f=
P
y
O
d
D
S
1

2

f
l
n = (n + kt)
0
0
sin D
y
n kt
f
\=
+
( y -coordinates of central maximum).
26. (b)
2
0
– sin
()
dy kD
dt
n kt
f
=
+
= velocity of central maximum.
27. (d) For central maxima to be formed at O
' –1 sin
'
n
n bd
n
æö
=f
ç÷
èø
Page 4

1. (d)
1
2
I 9
I1
=
2
1
2
2 max
min 1
2
I
1
I I 91
I I 91
1
I
éù
+
êú
éù
+
êú
==
êú
êú
-
ëû
êú-
êú
ëû
2
max
2
min
I 44
I1
2
==
2. (a)a
1
= 6 units, a
2
= 8 units
2
2
1
2 max
22
min
1
2
a
6
1
1
a I 8
I
6
a
1
1
8
a
éù
éù
+
+ êú
êú
ë û ëû
==
é ù éù
-
-
ê ú êú
ëû
ëû
max
min
I 49
I1
=
3. (b) The separation between the sucessive bright fringes
is-
b =
9
3
D 1 600 10
d
0.1 10
-
-
l ´´
=
´
b = 6.0 mm.
4. (b) w
a
= l/d
\ w
a
µ l  Þ
a water
a
() w
w
=
water
l
l
Þ
a water
a
() w
w
=
water
l
ml
Þ (w
a
)
water
= 0.15°
5. (b) I' = I
1
+ I
2
+ 2
12
I I cosf
I
1
= I, I
2
= 9I, f = p
I' = I + 9I + 2
2
9I cos p = 10I – 6I = 4I
6. (a) d =
Dl
b
.... (1)
According to quesion.
l = 5100 × 10
-10
m
b =
2
10
× 10
–2
m .... (2)
D = 2m,  d = ?.
From eqs. (1) and (2)
d =
8
3
2 51 10
2 10
-
-
´´
´
= 5.1 × 10
-4
m
7. (b) V =
max
max min min
max
max min
min
I
1
II I
I
II
1
I
-
-
=
+
+
.... (1)
2
1
2 max
min 1
2
I
1
I I
I I
1
I
æö
+
ç÷
ç÷
=
ç÷
ç÷ -
ç÷
èø
.... (2)
According to question
1
2
I 1
I4
=
.... (3)
From eqs. (2) and (3)
2
m ax
m in
1 9
1
I
4 4
9
1
I
1
1
4
4
æö
+
ç÷
ç÷
= ==
ç÷
-
ç÷
èø
.... (4)
From eqs. (1) and (4)
V =
[9 1]
[9 1]
-
+
=
8
10
= 0.8
8. (c) l =
6
( 1)t (1.6 1) 7 10
n7
-
m- - ´´
=
l = 6 × 10
–7
meter Þ l = 6000 Å
9. (d) b µ
1
d
\ d  On increasing d three times
b will become 1/3 times.
10. (b) \ PR = d Þ PO = d secq and CO = PO cos 2q
= d sec q cos 2q is
Q
O
C
A
P
B
R
q q
q
DPP/ P 52
146
Path difference between the two rays
D = CO + PO = (d sec q + d secq cos 2q)
Phase difference between the two rays is
f = p (One is reflected, while another is direct)
Therefore condition for constructive interference
should be
D =
3
,
22
ll
.......
or d sec q(1 + cos 2q) =
2
l
or ( )
2
d
2 cos
cos
q
q
=
2
l
Þ cos q =
4d
l
11. (b) n
1
l
1
= n
2
l
2
10 × 7000 = n
2
× 5000
n
2
= 14
12. (a, c) Path difference between the rays reaching infront of
slit S
1
is.
S
1
P – S
2
P = (b
2
+ d
2
)
1/2
– d
For distructive interference at P
b
S
1
S
2
d
P
S
1
P – S
2
P =
( ) 21
2
n-l
i.e., (b
2
+ d
2
)
1/2
– d =
( ) 21
2
n-l
Þ
( )
1/2
2
2
2n1
b
d1d
2
d
æö
-l
+ -= ç÷
ç÷
èø
Þ
( )
2
2
2n1 b
d 1 ...... d
2
2d
æö
-l
+ + -= ç÷
ç÷
èø
(Binomial Expansion)
Þ
( ) 2n1 b
2d2
-l
=
Þ =
( )
2
b
2n 1d -
For n = 1,2 ............, l =
22
bb
,
d 3d
13. (b) Distance of mth bright fringe from central fringes is
X
m
=
mD
d
l
14. (a, b) For microwave l =
c
f
=
8
6
3×10
10
= 300 m
d
Y
D
P
q
S
1
S
2
q
Dx
As Dx = d sin q
Phase difference f =
2p
l
(Path difference)
= ( )
2
d sin
p
q
l
=
2
300
p
(150 sin q) = p sin q
I
R
= I
1
+ I
2
+
12
2 I I cosq
Here I
1
= I
2
and f = p sin q
\ I
R
= 2I
1
[(p sin q)] = 4I
1
cos
2
sin
2
pq æö
ç÷
èø
I
R
will be maximum when cos
2

sin
2
pq æö
ç÷
èø
= 1
\ (I
R
)
max
= 4I
1
= I
o
Hence I = I
o
cos
2

sin
2
pq æö
ç÷
èø
If q = 0, then I = I
o
cosq = Io
If q = 30°, then I = I
o
cos
2
(p/4) = I
o
/2
If q = 90°, then I = I
o
cos
2
(p/2) = 0
15. (b) I = R
2
= a
1
2
+ a
2
2
+ 2a
1
a
2
cos d
=  I + 4I + 4I cos
2
p
= 5I
16. (b) Suppose the amplitude of the light wave coming from
the narrower slit is A and that coming from the wider
slit is 2A. The maximum intensity occurs at a place
where constructive interference takes place. Then the
resultant amplitude is the sum of the individual
amplitudes.
Thus,
A
max
= 2A + A = 3A
The minimum intensity occurs at a place where
destructive interference takes place. The resultant
amplitude is then difference of the individual
amplitudes.
Thus, A
min
= 2A – A = A.
\
2 2
max max
22
min
min
I (A) (3A)
9
I
(A ) (A)
= ==
DPP/ P 52
147
17. (b)
1
2
I 2
I1
=
11
22
aI 2
a I1
==
At the point of constructive interference, the resultant
amplitude becomes (a
1
+ a
2
) = 2 + 1 at the point of
destructive interference, the resultant amplitude is
(a
1
– a
2
) = 2 – 1
\
2 2
max 12
22
min
12
I (a a) ( 2 1)
34
I
(a a ) ( 2 1)
+ +
= ==
--
18. (d) For destructive interference :
Path difference= S
1
P – S
2
P = (2n –1) l/2.
For  n = 1, S
1
P – S
2
P= (2 × 1 – 1) l/2 = l/2.
n =2, S
1
P– S
2
P = (2 × 2 – 1) l/2 = 3l/2.
n = 3, S
1
P – S
2
P = (2× 3–1) l/2  = 5l/2.
n = 4, S
1
P – S
2
P = (2 × 4 – 1) l/2  = 7l/2.
n = 5,S
1
P – S
2
P = (2 × 5 – 1) l/2= 9l/2.
n =6,S
1
P – S
2
P = (2× 6– 1) l/2  = 11l/2.
So, destructive pattern is possible only for path
difference = 11l/2.
19. (b) The distance of a bright fringe from zero order fringe is
given by-
X
n
=
n D
d
l
D & d is constant
n
1
l
1
= n
2
l
2
n
1
= 16, l
1
= 6000 Aº, l
2
= 4800 Å
n
2
=
11
2
n 16 6000
4800
l ´
=
l
= 20
20. (c)n
1
l
1
= n
2
l
2
for bright fringe
n (7.5 × 10
–5
) = (n + 1) (5 × 10
–5
)
n =
5
5
5.0 10
2.5 10
-
-
´
´
= 2.
21. (b) X
n
=
nD
d
l
or X
3
=
3D
d
l
X
3
=
8
3 (5000 10 cm) 200cm
0.02cm
-
´ ´´
= 1.5 cm.
22. (b, d)
max
max
I
9
I
=
Þ
2
12
12
a + a
a – a
æö
ç÷
èø
= 9  Þ
12
12
a + a
a – a
= 3
Þ
1
2
a
a
=
3 + 1
3 – 1
Þ
1
2
a
a
= 5 There I
1
: I
2
= 4 : 1
23. (a) Fringe width b =
D
d
l
According to de Broglie,
Wavelength l =
hh
p 2meV
=
As V increases, l decreases, b decreases.
Also
1
andD
d
bµ bµ .
24. (a) n =
( 1)tD
d
m-
but b =
D
d
l
Þ
D
d
b
=
l
n = (m – 1) t b/l
20b = (m – 1) 2.5 × 10
–3
{ b/5000 × 10
–8
}
m – 1 =
8
3
20 5000 10
2.5 10
-
-
´´
´
= 0.4
m = 1.4.
25. (a) S
1
P – S
2
P
dy
D
=
For central maxima,
Dx = (n
0
+ kt) –sin0
dy
d
D
f=
P
y
O
d
D
S
1

2

f
l
n = (n + kt)
0
0
sin D
y
n kt
f
\=
+
( y -coordinates of central maximum).
26. (b)
2
0
– sin
()
dy kD
dt
n kt
f
=
+
= velocity of central maximum.
27. (d) For central maxima to be formed at O
' –1 sin
'
n
n bd
n
æö
=f
ç÷
èø
DPP/ P 52
148
Here n' = n
0
+ kt, n = refractive index of plate.
0
sin d
n n kt
b
f
= ++
28. (d) When d is  negligibly small, fringe width b which is
proportional to 1/d may become too large. Even a single
fringe may occupy the whole screen. Hence the pattern
cannot be detected.
29. (d) In Y oung's experiments fringe width for dark and white
fringes are same while in Y oung's double slit experiment
when a white light as a source is used, the central fringe
is white around which few coloured fringes are
observed on either side.
30. (d) When one of slits is covered with cellophane paper,
the intensity of light emerging from the slit is decreased
(because this medium is translucent). Now the two
interfering beam have different intensities or
amplitudes. Hence intensity at minima will not be zero
and fringes will become indistinct.
```

## Physics Class 12

105 videos|425 docs|114 tests

## Physics Class 12

105 videos|425 docs|114 tests

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